Let $X_1$ and $X_2$ be random variables (not of the same distribution and not independent). Both have a zero probability of being below $-1$. Their joint density is $\rho(x_1,x_2)$. Also, they both have finite expectations.

1 Answer
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It does, because $f$ is strictly concave. Indeed, consider its restriction to a line through $(t_1,t_2)$ in the direction $(v_1,v_2)$. In terms of parameter $s$ this restriction is
$$f(t_1+sv_1,t_2+sv_2)=\int_{-1}^\infty\int_{-1}^\infty \log(1+t_1x_1+t_2x_2+s(v_1x_1+v_2x_2))\rho(x_1,x_2)\,dx_1dx_2$$
The one-variable function $s\mapsto \log(1+t_1x_1+t_2x_2+s(v_1x_1+v_2x_2))$ is strictly concave unless $v_1x_1+v_2x_2=0$. Assuming the joint density is indeed a density and not some singular measure, we conclude that for $\rho$-a.e. $(x_1,x_2)$ strict concavity holds. Thus, integration against $\rho$ gives a strictly concave function.

If you want a more explicit argument, take the second derivative in $s$:
$$\frac{d^2}{ds^2}f(t_1+sv_1,t_2+sv_2)\\=-\int_{-1}^\infty\int_{-1}^\infty (v_1x_1+v_2x_2)^2 (1+t_1x_1+t_2x_2+s(v_1x_1+v_2x_2))^{-2}\rho(x_1,x_2)\,dx_1dx_2<0$$